Problem: What is the area of the region between the graphs of $f(x)=x^2-4x+1$ and $g(x)=3x-5$ from $x=1$ to $x=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{27}{2}$ (Choice B) B $4$ (Choice C) C $\dfrac{81}{2}$ (Choice D) D $\dfrac{125}{6}$
Solution: Visualizing the area We sketch the graphs of $~f~$ and $~g~$ first. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between $x=1$ and $x=4$. From this we are looking to evaluate: $ \int_{1}^{4}\left( g(x)-f(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{1}^{4} \left( 3x-5- (x^2-4x+1)\right) \,dx \\\\ &= \int_{1}^{4} \left( -x^2+7x-6\right) \,dx \\\\ &= -\dfrac{x^3}{3}+\dfrac{7x^2}{2}-6x~\Bigg|_{1}^{4} \\\\ &= \left( -\dfrac{64}{3} + 56 -24 \right) -\left( -\dfrac{1}{3} + \dfrac{7}{2} - 6 \right)\\\\ &= \dfrac{27}{2} \end{aligned}$ Answer The area is $\dfrac{27}{2}$ square units.